The closure of the Hurwitz matrices

Let $\mathcal{K}=\{A:\operatorname{Re}\lambda\le0\ \forall\lambda\in\operatorname{spec}A\}$.

"$\overline{\mathcal{H}}\subseteq\mathcal{K}$". Let $A_k\in\mathcal{H}$, $A_k\to A$, and let $\mu$ be an eigenvalue of $A$. We show that $\mu$ is a limit of eigenvalues of $A_k$. As in the density section: $\prod_i|\mu-\lambda_i(A_k)|=|\chi_{A_k}(\mu)|\to|\chi_A(\mu)|=0$, so $\min_i|\mu-\lambda_i(A_k)|\to0$. Pick $\lambda^{(k)}$ an eigenvalue of $A_k$ with $\lambda^{(k)}\to\mu$. Since $\operatorname{Re}\lambda^{(k)}<0$, passing to the limit gives $\operatorname{Re}\mu\le0$. So $A\in\mathcal{K}$.

"$\mathcal{K}\subseteq\overline{\mathcal{H}}$". Let $A\in\mathcal{K}$ and $\varepsilon>0$. The eigenvalues of $A-\varepsilon I$ are $\lambda-\varepsilon$ with $\lambda\in\operatorname{spec}A$, and $\operatorname{Re}(\lambda-\varepsilon)\le-\varepsilon<0$. So $A-\varepsilon I\in\mathcal{H}$, and $A-\varepsilon I\to A$ as $\varepsilon\to0$.

Remark. $\mathcal{H}$ is open (continuity of eigenvalues), and $\mathcal{K}$ is closed; the boundary consists of matrices in $\mathcal{K}$ with an eigenvalue on the imaginary axis. $e^{tA}\to0$ for $A\in\mathcal{H}$: this is the stability of $x'=Ax$. Lesson. Shifting by $\varepsilon I$ moves the whole spectrum; it is the simplest perturbation.